theorem
  V1 is linearly-closed & V2 is linearly-closed &
    V3 = {v + u : v in V1 & u in V2} implies
    V3 is linearly-closed
  proof
    assume that
    A1: V1 is linearly-closed & V2 is linearly-closed and
    A2: V3 = {v + u : v in V1 & u in V2};
    thus for v, u st v in V3 & u in V3 holds v + u in V3
    proof
      let v, u;
      assume that
      A3: v in V3 and
      A4: u in V3;
      consider v2, v1 such that
      A5: v = v1 + v2 and
      A6: v1 in V1 & v2 in V2 by A2,A3;
      consider u2, u1 such that
      A7: u = u1 + u2 and
      A8: u1 in V1 & u2 in V2 by A2,A4;
      A9: v + u = ((v1 + v2) + u1) + u2 by A5,A7,RLVECT_1:def 3
      .= ((v1 + u1) + v2) + u2 by RLVECT_1:def 3
      .= (v1 + u1) + (v2 + u2) by RLVECT_1:def 3;
      v1 + u1 in V1 & v2 + u2 in V2 by A1,A6,A8;
      hence thesis by A2,A9;
    end;
    let a, v;
    assume v in V3;
    then consider v2, v1 such that
    A10: v = v1 + v2 and
    A11: v1 in V1 & v2 in V2 by A2;
    A12: a * v = a * v1 + a * v2 by A10,VECTSP_1:def 14;
    a * v1 in V1 & a * v2 in V2 by A1,A11;
    hence thesis by A2,A12;
  end;
